Exchange limit and derivative: $\lim_{t\to\infty} f(t)=0$ implies $\lim_{t\to\infty} \frac{d f(t)}{dt}=0$?

Let $f(t)$ be absolutely continuous, uniformly bounded and such that $\lim_{t\to\infty} f(t)=0$. It should be true that $$\lim_{t\to\infty} \frac{d}{dt} f(t)=0$$ but how can one prove it?

I'm trying to find the formal justification that allows to exchange the limit and the derivative.

• I think your claim is false: consider $f(t) = \frac{1}{t} \sin (e^t)$ ($f(t) \to 0$ but $f'(t)$ is unbounded). – Connor Harris Mar 21 '17 at 15:40
• @ConnorHarris That function is not uniformly bounded. – Cameron Williams Mar 21 '17 at 15:43
• @ConnorHarris I agree that that function is uniformly bounded and $f(t)\to 0$. But I'm not sure that it is absolutely continuous. – user52227 Mar 21 '17 at 15:51
• Ah, you're right, it's not absolutely continuous. – Connor Harris Mar 21 '17 at 15:53
• @CameronWilliams Cameron, yes true. But inasmuch as $t\to \infty$, it doesn't matter much as to the behavior of $f$ except in a neighborhood of $\infty$. – Mark Viola Mar 21 '17 at 15:57

The result is actually false. Construct a continuous integrable function $g:\mathbb{R}\rightarrow\lbrack0,\infty)$ such that $$\liminf_{x\rightarrow\infty}g(x)=0<\limsup_{x\rightarrow\infty}g(x).$$ Note that this is done by taking $g$ to be a piecewise affine function which is zero most of the time and at every $n\in\mathbb{N}$, $g(n)=1$ and the area is the one of a triangle of height one and basis $\frac{1}{n^{2}}$. Then define $$f(x)=\int_{x}^{\infty}g(t)\,dt.$$ The function $f$ is absolutely continuous since $g$ is integrable, $\lim_{x\rightarrow\infty}f(x)=0$ and $f$ is bounded by $\int_{\mathbb{R}% }g(t)\,dt$ but $f^{\prime}(x)=-g(x)$ does not go to zero.
• Thanks. $g(x)$ does not go to zero because the limit of $g(x)$ does not exist, am I correct? If the limit would exist then we may be in a position to apply barbalat’s lemma to conclude that the result is true mathpost.asu.edu/~hliu/Things2Explain/… – user52227 Mar 21 '17 at 17:13
• correct, if the limit existed, it would have to be zero. For an integrable function you always need $\liminf_{x\to\infty}|g(x)|=0$ – Gio67 Mar 21 '17 at 17:17